Question: Find one value of $x$ that is a solution to the equation: $(x^2+1)^2-5x^2-5=0$ $x=$
We could solve for $x$ by expanding $(x^2+1)^2$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a shorter and more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Note that $-5x^2-5=-5({x^2+1})$. This means that we can rewrite the equation as: $({x^2+1})^2-5({x^2+1})=0$ If we let ${p}={x^2+1}$, we can see that this equation is in the form: ${p}^2-5{p}=0$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2-5{p}&=0\\\\ {p}({p}-5)&=0\\\\ {p}=0\ &\text{or} \ \ {p}=5 \end{aligned}$ Since ${p}={x^2+1}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${x^2+1}=0\ \ \ \text{or} \ \ \ {x^2+1}=5$ Note that there are no real solutions to the equation ${x^2+1}=0$. [Why not?] When we solve $x^2+1=5$, we find that $x=\pm{2}$. In conclusion, the two solutions of the equation $(x^2+1)^2-5x^2-5=0$ are $x=2$ and $x=-2$. [Is there another way to solve for x?]